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# Sin x / 1 - cos x + sin x / 1 + cos x = 2 csc x Prove the identity

$RTP: \frac{sin(x)}{1 - cos(x)} + \frac{sin(x)}{1 + cos(x)} = 2cosec(x)$ Important identities to know: $sin^{2}(x) + cos^{2}(x) = 1$ This can be rearranged to make either functions the subject: $sin^{2}(x) = 1 - cos^{2}(x)$ $cos^{2}(x) = 1 - sin^{2}(x)$ Thus, $LHS = \frac{sinx}{1 - cos(x)} + \frac{sin(x)}{1 + cos(x)}$ By simplifying into one fraction: $LHS = \frac{sin(x)(1 + cos(x)) + sin(x)(1 - cos(x))}{(1 - cos(x))(1 + cos(x)}$ $=\frac{sin(x)[1 + cos(x) + 1 - cos(x)]}{1 - cos^{2}(x)}$ $=\frac{2sin(x)}{sin^{2}(x)}$ $=\frac{2}{sin(x)}$ Now, $\frac{1}{sin(x)} = cosec(x)$ Thus, we can say: $LHS = 2cosec(x) = RHS$ (as required)